Question

What force is applied on a piston of area of cross section 2cm² to obtain a force 150 N on the piston of area of cross section 12cm² in a hydraulic machine.

give full answer with statements
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Answer 1

Explanation:

pressure 1 = pressure 2 (Pascal's law)

F1/A1 = F2/A2 ( pressure = force/ area )

(conversion of units to si may be ignored as it will eventually cancel since we have equated same physical quantities)

x/ 2= 150 /12

x = 150÷6 = 25 N

Answer 2

By the principle of hydraulic machine,

Pressure on smaller piston = pressure on wider piston.

.・. P1 = P2

$\frac{F1}{A1} = \frac{F2}{A2} \\ \\ \implies\frac{F1}{2 \times {10}^{ - 4} } = \frac{150}{12 \times {10}^{ - 4} } \\ \\ \implies \: F1 \: = \frac{150}{12 \times {10}^{ - 4} } \times 2 \times {10}^{ - 4} \\ \\ \implies \: F1 \: = \frac{ \cancel{150} \: \: ²⁵ }{ \cancel{12 }\times \cancel{ {10}^{ - 4} }} \times \cancel{2 }\times \cancel{{10}^{ - 4} } \\ \\ \ = 25 \: N....... \: \: \: (ans)$